anonymous
  • anonymous
A 2.0 µC point charge travels a distance of 0.20 cm between two parallel charged plates from the negative end toward the positive end. The electric field has a strength of 500.0 N.C. What is the change in voltage? (µC=1.0*10^-6 C)
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
A. -100 V B. +1.0 V C. 0 V D. +0.20 mV
Michele_Laino
  • Michele_Laino
the subsequent relationship holds: \[\Delta V = Ed\] |dw:1433273196060:dw|
anonymous
  • anonymous
ok! so what happens next? :/

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anonymous
  • anonymous
500.0 x 0.2
anonymous
  • anonymous
=100
anonymous
  • anonymous
+1.0 V is your answer! :)
anonymous
  • anonymous
ohh okay!! thank you!!:D
Michele_Laino
  • Michele_Laino
no, it is: \[\Large \Delta V = Ed = 500 \times 2 \times {10^{ - 3}} = ...\] since the distance is 0.2 cm = 2*10^(-3) meters
anonymous
  • anonymous
so +1.0V is the solution?
Michele_Laino
  • Michele_Laino
yes! since our charge is approaching to the positive plane, so its potential is increasing, namlely final potential is larger than initial potential so the difference between them is positive
anonymous
  • anonymous
yay! thank you both!!
Michele_Laino
  • Michele_Laino
:)

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